Selim Ghazouani
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 Jan21 revised List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$. added 10 characters in body Jan21 revised List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$. added 9 characters in body Jan21 comment List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$. Sorry I should have precised I am only interested in Lie subgroups of real dimension at least one. Jan21 asked List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$. Dec30 asked Difference between diffeomorphisms fixing a point or a whole neighborhood. Dec29 accepted Exact sequences and (semi) direct product Dec29 comment Exact sequences and (semi) direct product Thanks. A hint for point 3. ? Dec29 asked Exact sequences and (semi) direct product Dec25 comment Geodesic flow on a manifold with negative curvature is ergodic If you are interested in the proof of Mostow rigidity, you can certainly avoid using this result. Dec19 revised Linear group action over an hermitian space. added 453 characters in body Dec19 asked Linear group action over an hermitian space. Dec14 comment Closed but not exact one-form on $S^2$ Ok I read a bit fast your hypothesis and missed the fact that you removed 3 points. Whatever, I'm not sure I understand what you're asking for in your comment. Could you precise a bit ? Dec14 comment Closed but not exact one-form on $S^2$ This cannot happen on $S^2$ because the non-exactness of a closed form must come from a hole on the surface (in rough words). You should try to prove the following : let $S$ be a surface which is simply connected(this express the fact that $S$ has no hole), then every closed $1$-form is exact. Dec14 comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite? Other remark, such a matrix will always be of rank at most $1$, since all lines are colinear to $(f(x_1),...f(x_n))$. It implies that it will never be definite positive provided that $n \geq 2$. Dec14 comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite? First of all, put $f \equiv 0$ to get a matrix which is not positive-definite since $\overline{K} = 0$ . Dec11 comment Let $A$ be an abelian group. Show that $\mathrm{Hom}(\mathbb Z, A)$ is isomorphic to $A$. Assuming $Z$ is the set of integers, I invite you to check that $\varphi \in \mathrm{Hom}(\mathbb{Z}, A) \longmapsto \varphi(1)$ is the isomorphism you are seeking. Dec11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? I'm not sure. Here you are using the fact that complex eigenvalues are conjugates, and for each pair of conjugate you use only one eigenvector and take the real and imaginary part. DO you see what I mean ? Dec11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Ok I understand how you want to proceed but now we should check that $e_i,x_j,y_j$ actually form a basis for $\mathbb{R}^n$. I think that using computation you used to prove that $x_j$ and $y_j$ are linearly independent should work but I doesn't seem obvious. Dec11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Actually I don't really understand how you find the $f_j$ and $g_j$. Precisely, why do their coefficients belong to $\mathbb{R}$ since the diagonalization ii over $\mathbb{C}$ ? Dec11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Yes exactly. I'm not sure how to be clearer :)